We will present only the case \(b\in BMO^{-1}(\mathbb {R}^2)\)  since the case \(u\in BMO^{-1}(\mathbb {R}^2)\)  can be dealt with in a similar manner.
Recall that the Riesz transform of a function \(f\)  is defined as \(Rf := \nabla ((-\Delta )^{\frac{1}{2}}f\) . Because the Riesz transform is a bounded operator from \(BMO(\mathbb {R}^n)\)  to \(BMO(\mathbb {R}^n)\)  (see for instance [1]}), we have that the stream function \(\psi \)  associated to \(b\)  belongs to \(BMO(\mathbb {R}^2)\) .
Recall our cut-off function \(\varphi \in C^{\infty }_0(\mathbb {R}^2)\)  such that \(0\le \varphi \le 1\)  with \(\varphi \equiv 1\)  in \(B(1/2)\)  and \(\varphi \equiv 0\)  in \(B\setminus B(3/4)\) . As before, for \(R>1\) , set \(\varphi _R(x) := \varphi (x/R)\) . We get from (REF ) that
\(c_0\int _{B(R)} \varphi _R \mathrm {\,d}x = -\int _{B(R)}\Delta \psi \varphi _R \mathrm {\,d}x + \int _{B(R)} u\cdot \nabla (\psi - [\psi ]_{,R})\varphi _R \mathrm {\,d}x.\) 
This implies,
\(|c_0|R^2\int _B\varphi \mathrm {\,d}x &\le R\Vert \nabla b\Vert _{L_2}\Vert \varphi \Vert _{L_2} + \left| \int _{B(R)}(\psi - [\psi ]_{,R}) u\cdot \nabla \varphi _R \mathrm {\,d}x \right|\\&\le c(\varphi )\left(R\Vert \nabla b\Vert _{L_2(\mathbb {R}^2)} + \Vert \psi \Vert _{BMO(\mathbb {R}^2)} \Vert u\Vert _{L_2(B(R))}\right).\) 

From Lemma REF , we get that
\(|c_0| \le c(\varphi ,b,u)\left( \frac{1}{R} + \frac{\sqrt{\log {2R}}}{R}\right) \rightarrow 0\mbox{ as }R\rightarrow \infty .\) 
This concludes the proof.
